Step-by-step Solution

1. Identify Given Information and Define Variables Let the weak monobasic acid be HA. We are given:

   • Limiting molar conductivity at 298 K, $\Lambda_m^\circ = 4 \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$.
   • Initial State (before dilution):
     • Let the initial concentration be $C_1 = C$.
     • Molar conductivity, $\Lambda_{m,1} = y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$.
     • Degree of dissociation is $\alpha_1 = \alpha$.
   • Final State (after 20 times dilution):
     • The new concentration is $C_2 = C_1 / 20 = C / 20$.
     • The new molar conductivity is $\Lambda_{m,2} = 3y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}$.
     • Let the new degree of dissociation be $\alpha_2$.

2. Calculate the Degree of Dissociation (\\[alpha]) The degree of dissociation for a weak electrolyte is given by the ratio of its molar conductivity at a certain concentration to its limiting molar conductivity: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

   • For the initial state: $\alpha_1 = \frac{\Lambda_{m,1}}{\Lambda_m^\circ} = \frac{y \times 10^2}{4 \times 10^2} = \frac{y}{4}$
   • For the final state (after dilution): $\alpha_2 = \frac{\Lambda_{m,2}}{\Lambda_m^\circ} = \frac{3y \times 10^2}{4 \times 10^2} = \frac{3y}{4}$

3. Apply Ostwald's Dilution Law For a weak monobasic acid, the dissociation constant ($K_a$) is given by Ostwald's dilution law: $K_a = \frac{C\alpha^2}{1-\alpha}$ The temperature remains constant at 298 K, so the value of $K_a$ does not change upon dilution.

   • For the initial state: $K_a = \frac{C_1 \alpha_1^2}{1 - \alpha_1} = \frac{C (y/4)^2}{1 - y/4} = \frac{C y^2 / 16}{(4-y)/4} = \frac{C y^2}{4(4-y)} \quad ... (i)$
   • For the final state: $K_a = \frac{C_2 \alpha_2^2}{1 - \alpha_2} = \frac{(C/20) (3y/4)^2}{1 - 3y/4} = \frac{(C/20) (9y^2/16)}{(4-3y)/4} = \frac{9 C y^2}{20 \times 16} \times \frac{4}{4-3y} = \frac{9 C y^2}{80(4-3y)} \quad ... (ii)$

4. Equate the Expressions for $K_a$ and Solve for y Since $K_a$ is constant, we can equate equations (i) and (ii): $\frac{C y^2}{4(4-y)} = \frac{9 C y^2}{80(4-3y)}$ Assuming $C \neq 0$ and $y \neq 0$, we can cancel the $C y^2$ term from both sides: $\frac{1}{4(4-y)} = \frac{9}{80(4-3y)}$ Cross-multiplying gives: $80(4-3y) = 9 \times 4 (4-y)$ $80(4-3y) = 36(4-y)$ Divide both sides by 4 to simplify: $20(4-3y) = 9(4-y)$ $80 - 60y = 36 - 9y$ Rearranging the terms to solve for $y$: $80 - 36 = 60y - 9y$ $44 = 51y$ $y = \frac{44}{51}$

5. Calculate the Numerical Value of y Calculating the value of the fraction: $y \approx 0.862745...$ Rounding to two decimal places, we get: $y = 0.86$

Thus, the value of y is 0.86.